18 Calculus of residues-II

TABLE OF CONTENTS

LEARNING OBJECTIVES

C a l c u l u s o f r e s i d u e s – I I

1. Evaluation of definite integrals

One of the first applications of the Cauchy’s residue theorem was made in the evaluation of definite integrals. The integrals that can be evaluated by the method of residues can also be evaluated by other more traditional methods. However in many cases the method of residues turns out to be much simpler to apply.

In the last unit we laid down the groundwork for the evaluation of definite integrals by the use of residue theorem.

We now take up the actual task of evaluating such integrals.

The integrals that can be evaluated by this method fall into four main categories:

Sometimes integrals can be transformed by a complex change of variable to a form which can then be evaluated by these methods.

2. Integrals of the type

The method is useful when the function f is a rational function of sin and cos . To evaluate the integral we make the substitution

Hence the integral becomes

The contour C is the unit circle |z| = 1 traversed in the positive (anticlockwise) direction.

Example-1

Let us evaluate the integral

Here a is real and greater than 1.

First of all we notice that the integral is invariant under , so that

Example-2

Evaluate the integral

3. Integrals of the type

We first consider integrals in which the function f(x) is a real valued rational function in the real variable x. We assume that the degree of the denominator is at least two more than that of the numerator and f has no poles on the real axis, so that the integral exists.

First of all we extend the definition of the function to the whole complex plane, so that we have function of a complex variable f(z). Now consider the integral over the closed contour C, which consists of the real axis from –R to R and the semicircle Γ in the upper half plane, as shown in the figure below:

Example

We expect the integral over the semicircle to vanish in the limit of large R. Let us verify it rigorously. On the semicircle

Hence the integral over the semicircle is indeed zero, so that

4. Integrals of the type

Example-1

Let us evaluate the integral

Example-2

By use of Jordan’s lemma, we can evaluate integrals in which the degree of the denominator is only one more that of the numerator. As an example, now evaluate the integral

Consider instead the integral

4.1 An alternative method- the rectangular contour

say. Then

Thus we see that I tends to zero as R tends to infinity and the result is proved.

4.2 Indenting of contour

Example

Let us evaluate the integral

Consider the function

This function has a pole at the origin which lies on the path of integration. One way out is to consider instead the function

which has a removable singularity at the origin. However we wish to study the method of indenting the contour. We use the contour shown above where the point c is the origin. The integral now consists of four parts:

Since there are no singularities inside the contour C, by Cauchy’s residue theorem

As we have already seen, on use of Jordan’s lemma, the integral on the part Γ tends to zero as R tends to infinity. Combining all this together, we obtain

On comparing real and imaginary parts we obtain

5. Integrals of the type

Let the function f(x) be a rational function which has either no poles or simple poles on the positive real axis. If further

There are three main methods to evaluate such integrals by the method of residues.

5.1 Method-1

Here the contour C consists of the real axis from –R to R and the usual semicircle in the upper half plane. If there are poles on the real axis, they are indented in the manner described above.

5.2 Method-2

The value of the intended integral is obtained by equating imaginary parts.

5.3 Method-3

Example

We illustrate all the three methods by evaluating the following integral

Method-1

Write the integral as

Consider therefore the integral

where C is the rectangular contour of Figure-4. The integrand has only one pole inside the contour, a simple pole at the point

The integral over the four parts of the contour is

Method-2

The contour consists of six parts; hence

Now

The principle value is with respect to the singularity at z = 1. On equating the imaginary part we obtain the desired integral

In addition, on comparing the real parts we get

The second method has provided the value of two definite integrals in one go.

Method-3

In this case the contour is as shown alongside. Inside the contour there is only one singularity, a simple pole at z = 1. The residue at the pole is

6. Integrals of the type

Here log refers to the principal value of the logarithm. The function f is a rational function of the real variable x. We assume that the degree of the denominator of f is at least two more than that of the numerator. We further assume that f has no poles on the nonnegative real axis, so that the integrals are convergent. We will also assume that f is an even function of x.

Let us consider the function